Open Question
A leaf of area 40cm² and mass 4.5 x 10⁻⁴ kg directly faces the Sun on a clear day. The leaf has an emissivity of 0.85 and a specific heat of 0.80 kcal/kg · K .
(c) Will the temperature rise continue for hours? Why or why not?
20. Heat and Temperature
Heat Transfer
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300g of water at 40°C is in an insulated cup. What mass of copper at 500°C must be added to heat the water to 60°C?
A
50g
B
100g
C
250g
D
300g
E
150g
F
200g
Verified step by step guidance1
Identify the principle of conservation of energy, which states that the heat lost by the copper will be equal to the heat gained by the water.
Use the formula for heat transfer: \( Q = mc\Delta T \), where \( Q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
Set up the equation for the water: \( Q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial, water}) \). Substitute \( m_{water} = 300 \text{ g} \), \( c_{water} = 4.18 \text{ J/g°C} \), \( T_{final} = 60°C \), and \( T_{initial, water} = 40°C \).
Set up the equation for the copper: \( Q_{copper} = m_{copper} \cdot c_{copper} \cdot (T_{final} - T_{initial, copper}) \). Substitute \( c_{copper} = 0.385 \text{ J/g°C} \), \( T_{final} = 60°C \), and \( T_{initial, copper} = 500°C \).
Equate the heat gained by the water to the heat lost by the copper: \( m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial, water}) = m_{copper} \cdot c_{copper} \cdot (T_{initial, copper} - T_{final}) \). Solve for \( m_{copper} \) to find the mass of copper needed.
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